Prove that a group of even order must have an element of order 2.

Step 1: Let $G$ be a group of even order, and let $|G|$ denote the order of $G$. By definition, the order of a group refers to the number of elements it contains.
Since $G$ has even order, we can write $|G|=2n$, where $n$ is a positive integer.
Step 2: Now, consider the set $S=\{g\in G:g\ne g^{-1}\}$. In other words, $S$ consists of all the elements of $G$ that are not equal to their own inverses.
Assume, for the sake of contradiction, that $S$ is empty. This would mean that every element in $G$ is its own inverse. Therefore, for any $g\in G$, we would have $g=g^{-1}$.
Next, let's consider the identity element $e$ of $G$. By the definition of an identity element, $e$ multiplied by any element $g\in G$ results in $g$ itself. Thus, $eg=g$.
Step 3: Now, since $G$ is a group, it follows that $g^{-1}\in G$ for any $g\in G$. Therefore, we can consider the product of an element and its inverse: $g{g}^{-1}$.
Using the fact that $g=g^{-1}$ for all $g\in G$ (assuming $S$ is empty), we can rewrite $g{g}^{-1}$ as $ge$.
This simplifies to $g$, since $ge=g$ for any $g\in G$.
Thus, we have shown that $g=g^{-1}$ for all $g\in G$, which contradicts the assumption that $S$ is empty.
Therefore, the set $S$ cannot be empty. Hence, there exists an element $h\in S$ such that $h\ne h^{-1}$.
Step 4: Since $h$ is not equal to its own inverse, we have $h^{-1}\ne h$. This means that the order of $h$ is greater than 1.
However, since $|G|=2n$, which is even, and $h$ has an order greater than 1, it follows that the order of $h$ must be even.
By Lagrange's theorem, the order of an element divides the order of the group. Since the order of $h$ is even and divides the order of $G$, it must be a divisor of $2n$.
The only possible divisors of $2n$ are 1, 2, $n$, and $2n$.
Since $h$ has an order greater than 1, we can conclude that the order of $h$ cannot be 1 or $2n$.
Therefore, the order of $h$ must be either 2 or $n$. However, since $h\ne h^{-1}$, it cannot have order $n$.
Hence, the order of $h$ must be 2. Therefore, we have proven that a group of even order must have an element of order 2.

To prove that a group of even order must have an element of order 2, we can proceed as follows:
Let $G$ be a group of even order. By the Lagrange's theorem, the order of any element in $G$ must divide the order of the group. Since the order of $G$ is even, it can be expressed as $|G|=2n$, where $n$ is a positive integer.
Now, suppose for contradiction that there exists no element in $G$ of order 2. That means, for every element $g$ in $G$, $|g|\ne 2$.
Consider the set $S=\{g,g^2:g\in G\}$. Since $|g|\ne 2$ for any $g$ in $G$, we can conclude that $|g^2|\ne 2$ as well. Therefore, each element in $S$ has order greater than 2.
Note that every element in $S$ is distinct. To see this, suppose $g^k=h^m$ for some $g,h\in G$ and positive integers $k$ and $m$. Then, raising both sides to the power of 2, we have $(g^k{)}^2=(h^m{)}^2$, which implies $g^{2k}=h^{2m}$. Since $g^{2k}$ and $h^{2m}$ have order greater than 2, we can conclude that $g^{2k}=h^{2m}$ if and only if $g^k=h^m$. Thus, the elements in $S$ are distinct.
Now, let's count the number of elements in $S$. Since $S$ consists of the elements $g$ and $g^2$ for every $g$ in $G$, we have $|S|=2|G|=4n$.
On the other hand, note that $S$ is a subset of $G$, and by Lagrange's theorem, the order of any subgroup of $G$ must divide the order of $G$. Therefore, the order of $S$ must divide $|G|$. However, we have $|S|=4n$, which does not divide $2n=|G|$. This leads to a contradiction.
Hence, our assumption that there exists no element of order 2 in $G$ is false. Therefore, a group of even order must have an element of order 2.